I have a variable that gets me the type Observable<BleHandler.BlePeripheral> after using flatMap on the array.
let scannedPeripheral: Observable<BleHandler.BlePeripheral> = instance.bleScan()
.flatMap{ Observable.from($0)}
But now I need to use that variable in another function that takes BleHandler.BlePeripheral:
instance.bleEstablishConnection(scannedPeripheral: scannedPeripheral)
Obviously it doesn't work. Is there a way to get my Observable<BleHandler.BlePeripheral> to just BleHandler.BlePeripheral
It depends on whether or not the function returns a value and what type of value it returns...
If the function is void and you are just calling it for side effects then:
let disposable = scannedPeripheral
.subscribe(onNext: { instance.bleEstablishConnection(scannedPeripheral: $0) })
If your function has side effects and returns an Observable then:
let returnValue = scannedPeripheral
.flatMap { instance.bleEstablishConnection(scannedPeripheral: $0) }
If the function has no side effects and you are just calling it to transform your value into a different value then:
let returnValue = scannedPeripheral
.map { instance.bleEstablishConnection(scannedPeripheral: $0) }
This last one is unlikely based on the name of the function, but I put it here for completeness.
I am confused with return in Swift. I understand it's used to return the value in a function, if used like this:
func double(value: int) -> Int {
return value * 2
}
But I often just see return being used, like in a guard statement in an optional binding like this:
guard let value = value else (
print ("nothing")
return
}
So what is the purpose of having just return in the guard statement like this? Actually, I often see this not only in guard statements when unwrapping optional values. I always find this problem when writing code, when I want to use an optional string from a dictionary.
let info = ["name": "sarah", "hometown": "sydney"]
class UserInfo {
func getTheName() -> String {
guard let name = info["name"] else { return }
return name
}
}
// Compile time error: "Non-void function should return a value"
I get this error even though I have written return name. Xcode still complains that I have not returned a value. Is it because of the return in the guard statement?
So, could you please tell me the purpose of return in Swift? It is confusing for me.
return without any argument returns Void. This form of the return statement can only be used with a function that returns Void.
Once the return statement executes, the function exits and no more code in your function executes. Since you have a return in the guard statement, the second return name won't be executed (it couldn't anyway since it wouldn't have a name to return), which is why you get a compiler error; the compiler looks at all of the paths that your function could take to return something and ensures that all of those paths return what the function signature says it will.
The function in your question states that it returns a String, so you can't simply say return in the guard statement as that returns Void, violating the contract expressed by your function signature.
You could return a default value that isn't Void:
func getTheName () -> String {
guard let name = info["name"] else {
return ""
}
return name
}
This could be written much more succinctly using the nil-coalescing operator; return info["name"] ?? ""
You can also use return in a function that returns Void (or has no explicit return type, in which case it is implicitly understood to return Void)
So you could have a function like:
func maybePrint(theMessage: String?) -> Void {
guard let msg = theMessage else {
return
}
print(msg)
}
You're on the right track.
In your guard statement inside getTheName(), the 'return' keyword will try to exit the function itself if the guard fails. But the function requires you to return a String and as such you get the compiler error.
Here is a portion of another SO answer to a similar question:
guard forces you to exit the scope using a control transfer statement.
There are 4 available to you:
return and throw both exit the function/method continue can be used
within loops (while/for/repeat-while) break can be used in loops
(while/for/repeat-while) to exit the immediate scope. Specifying a
label to break to will allow you to exit multiple scopes at once (e.g.
breaking out of nested loop structure). When using a label, break can
also be used in if scopes. Additionally, you may exit the scope by
calling a function that returns Never, such as fatalError.
Stack Overflow: If the Swift 'guard' statement must exit scope, what is the definition of scope?
I have a problem with some Swift 2 code, compiled using Xcode 7 beta 3.
I have a class (see below) that has an initializer that takes a function, f, which can throw. If f does not throw, then a member variable (self.result) should be set to an instance of an enum that wraps the value that f returns. If f does throw, then self.result should be set to an instance of the enum that indicates the value is absent. At the end of the initializer, self.result should not be nil. I have checked the case where the f does not throw, and the behavior is correct. However, in the case that f does throw, self.result is nil at the end of the initializer (the assert is triggered). If I single-step in the debugger, I see that self.result seems to instantaneously be set and then flashes back to being nil.
(Note: You might suggest I represent the absence of a result as nil, rather than wrapping in the enum. However, I need to model the scenario in which the result of f has not been computed yet, has been computed successfully, or an attemopt has been made to compute the result but it has failed. Hence the enum.)
Have I misunderstood how Swift 2's error handling works? Or, is the compiler/debugger etc. behaving incorrectly?
Thanks in advance.
internal enum Result<T> {
case Value(T)
case None
}
public final class MyClass<T> {
internal var result: Result<T>? = nil
private init(f: () throws -> T) {
let queueName = “some.string”
let queue = dispatch_queue_create(queueName, DISPATCH_QUEUE_CONCURRENT)
dispatch_async(queue) {
do {
let value = try f()
self.result = .Value(value)
}
catch {
self.result = .None
}
assert(self.result != nil, "Result must have value before block returns.")
}
}
}
In
self.result = .None
the left-hand side is an optional Result, therefore .None on
the right-hand side is inferred as Optional.None, and the statement
is equivalent to
self.result = nil
What you probably meant is
self.result = Result.None
and then the assertion does not fail anymore. Alternatively,
use a different enumeration value, e.g. case NoValue in your custom
type.
I'm using Xcode 6 Beta 4. I have this weird situation where I cannot figure out how to appropriately test for optionals.
If I have an optional xyz, is the correct way to test:
if (xyz) // Do something
or
if (xyz != nil) // Do something
The documents say to do it the first way, but I've found that sometimes, the second way is required, and doesn't generate a compiler error, but other times, the second way generates a compiler error.
My specific example is using the GData XML parser bridged to swift:
let xml = GDataXMLDocument(
XMLString: responseBody,
options: 0,
error: &xmlError);
if (xmlError != nil)
Here, if I just did:
if xmlError
it would always return true. However, if I do:
if (xmlError != nil)
then it works (as how it works in Objective-C).
Is there something with the GData XML and the way it treats optionals that I am missing?
In Xcode Beta 5, they no longer let you do:
var xyz : NSString?
if xyz {
// Do something using `xyz`.
}
This produces an error:
does not conform to protocol 'BooleanType.Protocol'
You have to use one of these forms:
if xyz != nil {
// Do something using `xyz`.
}
if let xy = xyz {
// Do something using `xy`.
}
To add to the other answers, instead of assigning to a differently named variable inside of an if condition:
var a: Int? = 5
if let b = a {
// do something
}
you can reuse the same variable name like this:
var a: Int? = 5
if let a = a {
// do something
}
This might help you avoid running out of creative variable names...
This takes advantage of variable shadowing that is supported in Swift.
Swift 3.0, 4.0
There are mainly two ways of checking optional for nil. Here are examples with comparison between them
1. if let
if let is the most basic way to check optional for nil. Other conditions can be appended to this nil check, separated by comma. The variable must not be nil to move for the next condition. If only nil check is required, remove extra conditions in the following code.
Other than that, if x is not nil, the if closure will be executed and x_val will be available inside. Otherwise the else closure is triggered.
if let x_val = x, x_val > 5 {
//x_val available on this scope
} else {
}
2. guard let
guard let can do similar things. It's main purpose is to make it logically more reasonable. It's like saying Make sure the variable is not nil, otherwise stop the function. guard let can also do extra condition checking as if let.
The differences are that the unwrapped value will be available on same scope as guard let, as shown in the comment below. This also leads to the point that in else closure, the program has to exit the current scope, by return, break, etc.
guard let x_val = x, x_val > 5 else {
return
}
//x_val available on this scope
One of the most direct ways to use optionals is the following:
Assuming xyz is of optional type, like Int? for example.
if let possXYZ = xyz {
// do something with possXYZ (the unwrapped value of xyz)
} else {
// do something now that we know xyz is .None
}
This way you can both test if xyz contains a value and if so, immediately work with that value.
With regards to your compiler error, the type UInt8 is not optional (note no '?') and therefore cannot be converted to nil. Make sure the variable you're working with is an optional before you treat it like one.
From swift programming guide
If Statements and Forced Unwrapping
You can use an if statement to find out whether an optional contains a
value. If an optional does have a value, it evaluates to true; if it
has no value at all, it evaluates to false.
So the best way to do this is
// swift > 3
if xyz != nil {}
and if you are using the xyz in if statement.Than you can unwrap xyz in if statement in constant variable .So you do not need to unwrap every place in if statement where xyz is used.
if let yourConstant = xyz {
//use youtConstant you do not need to unwrap `xyz`
}
This convention is suggested by apple and it will be followed by devlopers.
Although you must still either explicitly compare an optional with nil or use optional binding to additionally extract its value (i.e. optionals are not implicitly converted into Boolean values), it's worth noting that Swift 2 has added the guard statement to help avoid the pyramid of doom when working with multiple optional values.
In other words, your options now include explicitly checking for nil:
if xyz != nil {
// Do something with xyz
}
Optional binding:
if let xyz = xyz {
// Do something with xyz
// (Note that we can reuse the same variable name)
}
And guard statements:
guard let xyz = xyz else {
// Handle failure and then exit this code block
// e.g. by calling return, break, continue, or throw
return
}
// Do something with xyz, which is now guaranteed to be non-nil
Note how ordinary optional binding can lead to greater indentation when there is more than one optional value:
if let abc = abc {
if let xyz = xyz {
// Do something with abc and xyz
}
}
You can avoid this nesting with guard statements:
guard let abc = abc else {
// Handle failure and then exit this code block
return
}
guard let xyz = xyz else {
// Handle failure and then exit this code block
return
}
// Do something with abc and xyz
Swift 5 Protocol Extension
Here is an approach using protocol extension so that you can easily inline an optional nil check:
import Foundation
public extension Optional {
var isNil: Bool {
guard case Optional.none = self else {
return false
}
return true
}
var isSome: Bool {
return !self.isNil
}
}
Usage
var myValue: String?
if myValue.isNil {
// do something
}
if myValue.isSome {
// do something
}
One option that hasn't specifically been covered is using Swift's ignored value syntax:
if let _ = xyz {
// something that should only happen if xyz is not nil
}
I like this since checking for nil feels out of place in a modern language like Swift. I think the reason it feels out of place is that nil is basically a sentinel value. We've done away with sentinels pretty much everywhere else in modern programming so nil feels like it should go too.
Instead of if, ternary operator might come handy when you want to get a value based on whether something is nil:
func f(x: String?) -> String {
return x == nil ? "empty" : "non-empty"
}
Another approach besides using if or guard statements to do the optional binding is to extend Optional with:
extension Optional {
func ifValue(_ valueHandler: (Wrapped) -> Void) {
switch self {
case .some(let wrapped): valueHandler(wrapped)
default: break
}
}
}
ifValue receives a closure and calls it with the value as an argument when the optional is not nil. It is used this way:
var helloString: String? = "Hello, World!"
helloString.ifValue {
print($0) // prints "Hello, World!"
}
helloString = nil
helloString.ifValue {
print($0) // This code never runs
}
You should probably use an if or guard however as those are the most conventional (thus familiar) approaches used by Swift programmers.
Optional
Also you can use Nil-Coalescing Operator
The nil-coalescing operator (a ?? b) unwraps an optional a if it contains a value, or returns a default value b if a is nil. The expression a is always of an optional type. The expression b must match the type that is stored inside a.
let value = optionalValue ?? defaultValue
If optionalValue is nil, it automatically assigns value to defaultValue
Now you can do in swift the following thing which allows you to regain a little bit of the objective-c if nil else
if textfieldDate.text?.isEmpty ?? true {
}
var xyz : NSDictionary?
// case 1:
xyz = ["1":"one"]
// case 2: (empty dictionary)
xyz = NSDictionary()
// case 3: do nothing
if xyz { NSLog("xyz is not nil.") }
else { NSLog("xyz is nil.") }
This test worked as expected in all cases.
BTW, you do not need the brackets ().
If you have conditional and would like to unwrap and compare, how about taking advantage of the short-circuit evaluation of compound boolean expression as in
if xyz != nil && xyz! == "some non-nil value" {
}
Granted, this is not as readable as some of the other suggested posts, but gets the job done and somewhat succinct than the other suggested solutions.
If someone is also try to find to work with dictionaries and try to work with Optional(nil).
let example : [Int:Double?] = [2: 0.5]
let test = example[0]
You will end up with the type Double??.
To continue on your code, just use coalescing to get around it.
let example : [Int:Double?] = [2: 0.5]
let test = example[0] ?? nil
Now you just have Double?
This is totally logical, but I searched the wrong thing, maybe it helps someone else.
Since Swift 5.7:
if let xyz {
// Do something using `xyz` (`xyz` is not optional here)
} else {
// `xyz` was nil
}
I'm writing some Swift code where I have an array containing a generic type:
let _data: Array<T> = T[]()
Later in my code I need to determine the type stored in the array. I tried using the type casting technique described in the documentation (although it was not used for generics).
switch self._data {
case let doubleData as Array<Double>:
// Do something with doubleData
case let floatData as Array<Float>:
// Do something with floatData
default:
return nil // If the data type is unknown return nil
}
The above switch statement results in the following error upon compilation:
While emitting IR SIL function #_TFC19Adder_Example___Mac6Matrix9transposeUS_7Element__fGS0_Q__FT_GSqGS0_Q___
for 'transpose' at /code.viperscience/Adder/src/Adder
Library/Matrix.swift:45:3 :0: error: unable to execute
command: Segmentation fault: 11 :0: error: swift frontend
command failed due to signal (use -v to see invocation) Command
/Applications/Xcode6-Beta2.app/Contents/Developer/Toolchains/XcodeDefault.xctoolchain/usr/bin/swift
failed with exit code 254
Anybody know how I can cast my generic data to its actual type in order to take specific action?
In swift, as operator is something like dynamic_cast in C++, which can be used to down cast an object.
Say you have an object a of type A, and you can write let a as B only when type B is identical to type A, or B is a sub-class of A.
In your case, apparently Array<T> cannot always be down cast to Array<Double> or Array<Float>, so compiler reports errors.
A simple fix is to convert to AnyObject first, and then downcast to Array<Double> or Array<Float>:
let anyData: AnyObject = self._data;
switch anyData {
case let doubleData as? Array<Double>: // use as? operator, instead of as,
// to avoid runtime exception
// Do something with doubleData
case let floatData as? Array<Float>:
// Do something with floatData
default:
return nil // If the data type is unknown return nil
Suppose you have an array of buttons:
let views: [NSView] = [NSButton(), NSButton(), NSButton()]
You can use these casts:
let viewsAreButtons = views is [NSButton] // returns true
let buttonsForSure = views as! [NSButton] // crashes if you are wrong
let buttonsMaybe = views as? [NSButton] // optionally set
If you try to use as in a switch case like below, it will not work. The compiler (Swift 1.2 Xcode 6.3b1) says: "Downcast pattern of type [NSButton] cannot be used."
switch views {
case let buttons as [NSButton]:
println("Buttons")
default:
println("something else")
}
Call it a limitation. File a radar with your use case. The Swift team really seams to be listening for feedback. If you really want to get it to work, you can define your own pattern matching operator. In this case it would be something like this:
struct ButtonArray { }
let isButtonArray = ButtonArray()
func ~=(pattern: ButtonArray, value: [NSView]) -> Bool {
return value is [NSButton]
}
Then this works:
switch views {
case isButtonArray:
println("Buttons") // This gets printed.
default:
println("something else")
}
Try it in a Playground. Hope it helps!